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RMO Part-10!

1)Let A=\(\large(a^{2}+4ab+b^{2})\), where a and b are postive integers.

Prove that A≠2015. (HINT-Use Modular Arithmetic or Try to use the concept of Odd-Even numbers)

2)Prove that any prime number, \(\large(2^{2^n}+1)\) cannot be represented as a difference of 2 fifth powers of integers.(HINT-Expand the expression and You will get something in common)

3)Find all pairs (x,y) where x and y are integers such that \(\large(x^{3}+11^{3}=y^{3})\).(HINT-Try to apply identities)

4)If \(\large(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c})\)

Then Prove that,

\(\large(\frac{1}{a^{3}}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{a^{3}+b^{3}+c^{3}})\).(HINT-Factorize )

5)Prove that \(\large(2222^{5555}+5555^{2222})\) is divisible by 7.(HINT-Use Modular Arithmetic or Apply the concept of \(a^{n}±b^{n}\).)

Today I have posted some easy problems ,So I'm sure that you will be able to solve all these 5 problems in 1.5 hours.

Also try my Set RMO.

Note by Naitik Sanghavi
1 year, 8 months ago

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Here's my solution to (1).

Assume that ,

\(\large (a+b)^{2}+2ab=2015\)

\(\large (a+b)^{2}=2015-2ab\)

So, \(\large (a+b)^{2}\)=Odd

So,\(\large (a+b)\)=Odd(2q+1)For some positive integer q.

So,\(\large 2ab=4p\)For some integer p.

So, \(\large (2q+1)^{2}+4p=2015\)

\(\large 4(q^{2}+q+p)=2014\)

\(\large q^{2}+q+p=\frac{2014}{4}\)

Now we can see that R.H.S.is not an integer but In L.H.S. all are intgers .Thus it does not satisfy the given conditions.Hence Proved. Naitik Sanghavi · 1 year, 8 months ago

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4) let \(p(x) = x^3 + lx^2 + mx + p\) with roots a,b,c then \(p = mn\) and \(c = -b\) . Now it easy Dev Sharma · 1 year, 8 months ago

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@Dev Sharma Nice!!You got a different to solve this! Naitik Sanghavi · 1 year, 8 months ago

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5) \(5555=4 \pmod 7\) or \(5555^{2222}=4^{2222} \pmod 7\)

\(4^3=1 \pmod 7\) or \(4^{2220}=1 \pmod 7\) or \(4^{2222}=16=2 \pmod 7\)

Therefore, \(5555^{2222}=2 \pmod 7\).

Similarly,

\(2222=3 \pmod 7\) or \(2222^{5555}=3^{5555} \pmod 7\)

\(3^3=-1 \pmod 7\) or \(3^{5553}=(-1)^{1851}=-1=6 \pmod 7\) or \(3^{5555}=54=5 \pmod 7\)

Therefore, \(5555^{2222}=5 \pmod 7\)

Therefore \(5555^{2222}+2222^{5555}=2+5=7=0 \pmod 7\) Svatejas Shivakumar · 1 year, 8 months ago

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@Svatejas Shivakumar It seems that you have solved all problems of this note,You can my other problems of this set.Try my set Olympiad Preparation Set for an easier level !! Naitik Sanghavi · 1 year, 8 months ago

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2) Suppose \(2^{2^n}+1=a^5-b^5\),where \(a^5-b^5\) be the difference of two fifth powers.\(a^5-b^5=(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)\ \quad where \quad a>b\).

Since \(2^{2^n}+1 \quad is \quad prime, \quad a-b=1\).

By Fermat's Little theorem, \(2^{2^n}+1=a^5-b^5=a-b \pmod 5=1 \pmod 5\).

Thus \(2^{2^n}+1\) cannot be represented as a difference of two fifth powers.

Moreover, \(2^{2^n}+1\) cannot be represented as the difference of two pth powers, where p is an odd prime. Svatejas Shivakumar · 1 year, 8 months ago

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1) \(a^2 + 4ab + b^2 = 0, 2 or 1 mod4\) but 2015 = 3 mod4. A contradiction. Dev Sharma · 1 year, 8 months ago

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@Dev Sharma I seems that you have solved all problems on this Set,You can try my other problems of this set and Try my set Olympiad Preparation setfor an easier level!! Naitik Sanghavi · 1 year, 8 months ago

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@Dev Sharma Just a small correction, \(a^2+4ab+b^2\) can also be \(0 \pmod 4\).Try using \pmod for mod. I too did it exactly the same way. Svatejas Shivakumar · 1 year, 8 months ago

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@Svatejas Shivakumar Yes,of course!! You can post your solutions if you have solved any or have an different solution from others!!! Naitik Sanghavi · 1 year, 8 months ago

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@Dev Sharma Nice Observation!! I'm having an different solution!!You can try out other problems!!I will post my solutions after some days!!!!!!! Naitik Sanghavi · 1 year, 8 months ago

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3) Is (-11,0),(0,11) the only solutions? Svatejas Shivakumar · 1 year, 8 months ago

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@Svatejas Shivakumar Yes!!Correct but you need to prove that there are no other solutions too!! Naitik Sanghavi · 1 year, 8 months ago

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@Naitik Sanghavi @naitik sanghavi \(y= \sqrt [3]{x^3+11^3}=\sqrt [3]{(x+11)(x^2-11x+121)}\) which is only a perfect cube when either of the terms are zero. Hence, getting the above solution. Svatejas Shivakumar · 1 year, 8 months ago

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4) This one is quite long! After a lot of solving I got c=-a or c=-b . Substitute both the values and you get the second equation. Am I missing an easier method? Svatejas Shivakumar · 1 year, 8 months ago

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@Svatejas Shivakumar Here's my solution to (4). \(\large \frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)

\(\large \frac{a+b}{ab}=\frac{c-(a+b+c)}{(a+b+c)×c}\)

\(\large (a+b)×c×(a+b+c)=-(a+b)×ab\) (Don't remove (a+b) from both sides!Why?See Ahead!!!)

\(\large (a+b)×c×(a+b+c)+(a+b)ab=0\)

\(\large (a+b)[c×(a+c)+b×(a+c)]=0\)

\(\large (a+b)(b+c)(c+a)=0\)

a=-b or b=-c or c=-a.

Now substitute anyone value and it will proved!! Naitik Sanghavi · 1 year, 8 months ago

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@Naitik Sanghavi @naitik sanghavi Nice!!.I multiplied (ab+bc+ca) with (a+b+c),did a little bit of factoring and got \(c^2+(a+b)c+ab=0\) and finally solved the quadratic equation to get the above solution.Your solution is much faster upvoted!! Svatejas Shivakumar · 1 year, 8 months ago

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@Naitik Sanghavi You can also generalize it for powers of n and then by induction it will be proved for 3. Priyanshu Mishra · 1 year, 7 months ago

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Thanks for your contribution for the Brilliant RMO community :) Swapnil Das · 1 year, 8 months ago

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@Swapnil Das Its OK .I am also preparing for RMO so just posting some problems daily..!!!! Naitik Sanghavi · 1 year, 8 months ago

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